If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Differential equations>Unit 3

Lesson 2: Properties of the Laplace transform

# Laplace transform of the unit step function

Introduction to the unit step function and its Laplace Transform. Created by Sal Khan.

## Want to join the conversation?

• I keep getting stuck around 20 min where Sal explains that x is just a letter. I can't get my head around this since I thought that initially we made the substitution x = t - c
i.e. that f(t - c) = f(x)

Doesn't this mean that at the end we have to re-substitute t - c into the function such that we have the Laplace transform of the function f(t - c) factored by exp(-sc) as opposed to just the Laplace transform of the function f(t) factored by exp(-sc).

Thank you for any help :)
• Took me a while to understand this but I finally figured it out.

The issue here is in the normal way we write laplace transforms, the variable of integration is implied. This is because it doesn't normally matter, but it's biting us here.

The result Sal got was

e^(-sc) * laplace f(x) integrated with respect to x

This is exactly the same as
e^(-sc) * laplace f(t) integrated with respect to t
which is a property of integrals.

If you did substitute x = t - c, and wrote
e^(-sc) * laplace f(t-c)
you are actually writing
e^(-sc) * laplace f(t-c) integrated with respect to t
because we would assume the variable of integration is t.

However, to correctly substitute x = t - c you would need to write
e^(-sc) * laplace f(t-c) integrated with respect to t - c
This integral is the same as

e^(-sc) * laplace f(t) integrated with respect to t
which is what Sal concludes.

I hope this helps someone.
• Don't we have to assume that c > 0 for this? The bounds of integration in the original definition of the Laplace transform were from 0 to infinity. I understand that t values below c for the unit step function will cancel out the entire integral, but if we were to have a c value which was below 0 then wouldn't this in a sense violate our original definition of a Laplace transformation as the integral will give even more area under the curve? Or am I looking at this incorrectly?
• I think your point is what Sal mentioned in the beginning of the video. It's not that the unit step function requires the assumption c>0 in and of itself. Rather that for use with the Laplace transform, we are only interested in c values greater than zero since, as you mentioned, the transform is only concerned with t [0,inf).
• Defining the second unit step function, around 4min: when you say that you make the function jump up again to a y value of 2 at x=2π, the function also continues (and therefore is not a function) at y=0, unless you've somehow told it not to. The question then is, did you tell it not to, and if you did, how?
The unit step function (the u-something function of t) basically tells you where it is taking the step, in this case the first is at pi and the second at two-pi. The sign tells you which direction it is going, i.e. if it is negative it is going down and positive it is moving up. the coefficient in front, in this case they are both two's, tells you how far of a jump it is making. So in this function we start at a constant function of two, then when we have -2u[pi](t) for the next part we know when we get to pi on the t-axis it should decrease by 2, here equalling zero, and then when we have +2u[2pi](t) we know that after continuing at the zero from the previous part, when we get to two-pi we should increase by 2. I hope this clears it up a bit for you, let me know if you have any more questions. These guys can be a little tricksy sometimes.
• I have a question abt the video around 20.31 where SAL said that the laplace transformation of f(x) is almost equivalent to f(t). If x and t were independent variables, then i had no problem assuming that but here x and t are not independent variables because SAL makes a substitution earlier in the video that x = t-c , then wont the laplace transformation of f(x) would be equivalent to laplace transform of f(t-c). Please correct me where i m wrong?
• I had trouble wrapping my head around it too and so I substituted t back in and saw that it leads to the same conclusion. From what I understand, it's the presence of the unit step function (and that the entire function is 0 until t = c) that makes the Laplace transforms of f(x) and f(t) basically the same.

As a reminder,
t = x + c or x = t - c

Laplace{u_c(t) f(t-c)} = e^(-sc) * integral from x=0 to infinity of e^(-sx) f(x) dx

^Those equations were from around if that wasn't clear. Substituting back in t,

= e^(-sc) * integral from t=c to infinity of (e^(-s(t-c)) * u_c(t) * f(t-c) dt

If we were to subtract the bounds of the integral by c and replace t with t + c in the integrand, you get an equivalent integral. (Test this out yourself with a simpler integral if you find it difficult to swallow.) Also, infinity minus an arbitrary constant c is still infinity so the upper bound remains the same. So,

= e^(-sc) * integral from t=0 to infinity of e^(-s((t + c) - c)) * u_c(t + c) * f(t + c - c) dt

u_c (t + c) is always 1 within the interval [0, infinity) so we don't have to worry about it. Simplifying,

= e^(-sc) * integral from t=0 to infinity of e^(-st) * f(t) dt

That integral should look familiar enough to us as the Laplace transform of f(t). So,

Laplace{u_c(t) f(t-c)} = e^(-sc) * Laplace{f(t)}

• we assume that the term u_c(t) is 1 but this value seems to change for this function as it is peaks and troughs? or is he only referring to the y value of the function at the unit step?
• The graph is u_c(t) times f(t-c). It is the function f that is varying. u_c(t) merely multiplies f by 0 or 1.
• In this video Sal takes x as t and write f(x)=f(t) and say that both are just letters but what about that substitution x=t-c?
• wait I'm confused.. isn't this supposed to be the laplace transform of ONLY the step function?
The video is showing the laplace transform of step function u(t)*f(t-c)...
• Yes, the title of the video is slightly misleading, but if you want the laplace transform of just the unit step function, then you can just say f(t-c) = 1.
• i think it's also called the heavyside function, isn't it?